Math/CSE 1019C: Discrete Mathematics for Computer Science Fall 2012

1. Math/CSE 1019C:Discrete Mathematics for Computer ScienceFall 2012 Jessie Zhao jessie@cse.yorku.ca Course page: http://www.cse.yorku.ca/course/1019

2. Last date to drop courses without receiving a grade : Nov 9th • No Assignment is released today! • Test 2 on Nov 5th, • Ch2.1-2.5 • 7pm-8:20pm • Location: SLH F (Last name from A-L) and SLH A (Last name from M-Z) • Lecture: 8:40pm, SLH A. • You may pick up your Assignment 4 during my office hour: Nov 5th 2:00pm - 4:00pm (TEL 3056)

3. Mathematical Induction • Proof methods: direct proof, proof by cases, proof by contraposition, proof by contradiction, disproof by counterexample • Mathematical induction • very simple and powerful proof technique • often used to prove P(x) is true for all positive integers • “Guess” and verify strategy

4. Principle of Mathematical Induction (P(1) ⋀ ∀k(P(k)➝P(k+1)) ➝ ∀nP(n) • To prove that ∀nP(n), where n∈Z⁺ and P(n) is a propositional function, we complete two steps: • i) Basis step: Verify P(1) is true • ii)Inductive step: Show P(k)➝P(k+1) is true for arbitrary k∈Z⁺

5. Mathematical Induction • Knowing it is true for the first element means it must be true for the next element, i.e. the second element • Knowing it is true for the second element implies it is true for the third and so forth. • Need a starting point (Base case) P(1) P(k) → P(k+1) --------------------- , P(2) P(2) P(k) → P(k+1) --------------------- , … P(3)

6. How to show P(1) is true? • Replace n by 1 in P(n) • How to show P(k)➝P(k+1) is true? • Direct proof is normally used • (Inductive Hypothesis) Assume P(k) is true for some arbitrary k • Then show P(k+1) is true

7. Proving Summation • Example: Show that 1² + 2² + 3² + …+n² = n(n+1)(2n+1)/6, where n ∈ Z⁺ • Proof: • Basis case P(1): LHS=1² = 1, RHS=1*(1+1)(2*1+1)/6=1 • Inductive step: • Assume P(k) is true: 1² + 2² + 3² + …+k² = k(k+1)(2k+1)/6 • For P(k+1): Show that 1² + 2² + 3² + …+k²+ (k+1)²= (k+1)(k+2)(2k+3)/6 (Details on Board) • We showed P(k+1) is true under the assumption that P(k) is true. • By mathematical induction P(n) is true for all positive integers

8. Proving Inequalities • Example: n < 4n , where n ∈ Z⁺ • Proof • Basis case: P(1) holds since 1<4 • Inductive step: • Assume P(k) is true: k < 4k • For P(k+1): k+1 < 4k +1 < 4k + 4k = 2*4k < 4*4k = 4 k+1 • By mathematical induction P(n) is true for all positive integers

9. Proving divisibility • Example: Prove that n³-n is divisible by 3 whenever n is a positive integer. • Basis step: P(1): 1³-1=0, which is divisible by 3. • Inductive step: • Assume P(k) is true: k³ -k is divisible by 3 • For P(k+1): (k+1)³ -(k+1) =k³ +3k²+3k+1-k-1 =(k³-k ) +3(k²+k) From the assumption, k³-k is divisible by 3, so P(k+1) is true. By mathematical induction, P(n) is true for all positive integers

10. Mathematical Induction (Base case n≠1) • Basis case doe not have to be n=1 • How to show that P(n) is true for n=b, b+1, b+2,.... where b is an integer other than 1? • i) Basis step: Verify P(b) is true • ii)Inductive step: Show P(k)➝P(k+1) is true for arbitrary k∈Z

11. Show that 1+2+2²+...+2 n = 2 n+1 -1, where n∈N • Proof by induction: • Basis step: P(0): LHS=1, RHS = 21 -1 = 1. • Inductive step: • Assume P(k) is true for arbitrary k∈N, 1+2+...+2k = 2k+1 -1 • Need to show P(k+1): 1+2+....+2k +2 k+1 = 2 k+2 -1 is true. LHS = (1+2+...+2 k)+2 k+1 = 2 k +1-1+2 k+1 = 2 k+2-1 So LHS = RHS. We showed P(k+1) is true under the assumption that P(k) is true. • By mathematical induction P(n) is true for all natural numbers

12. Prove • Proof by induction: • P(n): a+ar+ar2.... +arn = (arn+1 -a)/(r-1) • Basis case: P(0): a = (ar-a)/(r-1). So, P(0) is true • Inductive step: • Assume P(k) is true for arbitrary k∈N, a+ar+ar2.... +ark = (ark+1 -a)/(r-1) • Need to show P(k+1): a+ar+ar2.... +ark + ark+1 = (ark+2 -a)/(r-1) is true. • LHS = (ar k+1-a)/(r-1) +ar k+1 = (ar k+1-a +ar k+2 -ark+1 )/(r-1) • So LHS = RHS. • We showed P(k+1) is true under the assumption that P(k) is true. • By mathematical induction P(n) is true for all natural numbers

13. A wrong proof, WHY? • Show that all horses are the same color. • Proof: • We do induction on the size of sets of horses of the same color. • Basis step: Obviously all sets of 0 horses (and all sets with 1 horse) are the same color • Inductive step: • Assume all horses are the same color for k horses • Now show it must be true for all sets of k+1 horses • Every set of k+1 horses has an overlap of horses which are the same color. • So k+1 horses have the same color. • Therefore all horses have the same color.

14. Strong Induction (P(1) ⋀ ∀k(P(1)⋀P(2)⋀...⋀P(k)➝P(k+1)))➝ ∀nP(n) • To prove that ∀nP(n), where n∈Z⁺ and P(n) is a propositional function, we complete two steps: • i) Basis step: Verify P(1) is true • ii)Inductive step: Show P(1)⋀P(2)⋀...⋀P(k)➝P(k+1)is true for arbitrary k∈Z⁺

15. Strong Induction Variation • A more general strong induction can handle cases where the inductive step is valid only for integers greater than a particular integer. • To prove that P(n) is true for all integer n≥b, we complete two steps: • i) Basis step: Verify P(b), P(b+1), ..., P(b+j) are true • ii)Inductive step: Show P(b+1)⋀P(b+2)⋀...⋀P(b+k)➝P(k+1) is true for every integer k≥b+j

16. Example: Show that if n is an integer greater than 1, then n can be written as the product of primes • Proof by strong induction: • First identify P(n), P(n): n can be written as the product of primes • Basis step: P(2): 2 is a prime number, so P(2) is true. • Inductive step: • Assume P(j) is true for 1<j≤k for an arbitrary k>1, i.e. j can be written as the product of primes when 1<j≤k • Need to show P(k+1). • Case 1: k+1 is prime, then P(k+1) is true. • Case 2: k+1 is composite k+1=a*b, where a and b are positive integers, and 2≤a,b≤k. By the assumption, a and b can be written as the product of primes. Then k+1 can be written as the product of primes.